# Electromagnetic Theory Questions and Answers – Relation of E,D,V

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This set of Electromagnetic Theory Multiple Choice Questions & Answers (MCQs) focuses on “Relation of E,D,V”.

1. The electric flux density and electric field intensity have which of the following relation?
a) Linear
b) Nonlinear
c) Inversely linear
d) Inversely nonlinear

Explanation: The electric flux density is directly proportional to electric field intensity. The proportionality constant is permittivity. D=ε E. It is clear that both are in linear relationship.

2. The electric field intensity is the negative gradient of the electric potential. State True/False.
a) True
b) False

Explanation: V = -∫E.dl is the integral form. On differentiating both sides, we get E = -Grad (V). Thus the electric field intensity is the negative gradient of the electric potential.

3. Find the electric potential for an electric field 3units at a distance of 2m.
a) 9
b) 4
c) 6
d) 3/2

Explanation: The electric field intensity is the ratio of electric potential to the distance. E = V/d. To get V = E X d = 3 X 2 = 6units.

4. Find the potential at a point (4, 3, -6) for the function V = 2x2y + 5z.
a) 96
b) 66
c) 30
d) -66

Explanation: The electric potential for the function V = 2x2y + 5z at the point (4, 3, -6) is given by V = 2(4)2(3) + 5(-6) = 96-30 = 66 units.

5. Find the electric flux density surrounding a material with field intensity of 2xyz placed in transformer oil ( εr = 2.2) at the point P(1,2,3) is
(in 10-10 units)
a) 2.1
b) 2.33
c) 2.5
d) 2.77

Explanation: D = εE, where ε = εo εr. The flux density is given by,
D = 8.854 X 10-12 X 2.2 X 2(1)(2)(3) = 2.33 X 10-10 units.

6. If potential V = 20/(x2 + y2). The electric field intensity for V is
40(x i + y j)/(x2 + y2)2. State True/False.
a) True
b) False

Explanation: E = -Grad (V) = -Grad(20/(x2 + y2)) = -(-40x i /(x2 + y2)2 – 40(y j)/(x2 + y2)2) = 40(x i + y j)/(x2 + y2)2. Thus the statement is true.

7. Find the potential of the function V = 60cos θ/r at the point P(3, 60, 25).
a) 20
b) 10
c) 30
d) 60

Explanation: Given V = 60cos θ/r. For r = 3m and θ = 60, we get V = 60cos 60/3 = 20cos 60 = 10 units.

8. Find the work done moving a charge 2C having potential V = 24volts is
a) 96
b) 24
c) 36
d) 48

Explanation: The work done is the product of charge and potential.
W = Q X V = 2 X 24 = 48 units.

9. If the potential is given by, V = 10sin θ cosφ/r, find the density at the point P(2, π/2, 0)
(in 10-12 units)
a) 13.25
b) 22.13
c) 26.31
d) 31.52

Explanation: Since V is given find out E.E = -Grad(V) = – Grad(10sin θ cosφ/r). From E, we can easily compute D. D = εE = 8.854 X 10-12 X 5/2 = 22.13 units.

10. If V = 2x2y + 20z – 4/(x2 + y2), find the density at A(6, -2.5, 3) in nC/m2.
a) 0.531i – 0.6373j – 0.177k
b) 0.6373i – 0.177j -0.531k
c) 0.177i – 0.6373j – 0.531k
d) 0.531i – 0.177j – 0.6373k

Explanation: Find E from V, E = -Grad (V). We get E at A(6,-2.5,3) as 59.97i – 71.98j -20k. Thus D = εE = 8.854 X 10-12 X
(59.97i – 71.98j -20k) = (0.531i – 0.6373j – 0.177k) nC/m2.

Sanfoundry Global Education & Learning Series – Electromagnetic Theory.
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